Question

Let $V_{1}$ be the volume of a given right circular cone with $O$ as the centre of the base and $A$ as its apex. Let $V_{2}$ be the maximum volume of the right circular cone inscribed in the given cone whose apex is $O$ and whose base is parallel to the base of the given cone. Then, the ratio $V_{2} / V_{1}$ is

• A. $\frac{3}{25}$
• B. $\frac{4}{9}$
• C. $\frac{4}{27}$
• D. $\frac{8}{27}$

Answer 1

Answer: (C)

$V_{1}$ = Volume of cone $ABC$
$H$ = Height of cone $V_{1}$
$R$ = Radius of cone $V_{1}$
$V_{2}$ = Volume of cone inscribed in $V_{1}$
$(H-h)$ = Height of cone $V_{2}$
r = Radius of cone $V_{2}$

$\Delta AO'Q$ and $\Delta\,AOC$ are similar
$\therefore \frac{AO'}{AO}=\frac{O'Q}{OC}$
$\Rightarrow \frac{h}{H} = \frac{r}{R}$
$\Rightarrow h = \frac{rH}{R}$
$\Rightarrow V_{2}=\frac{1}{3} \pi r^{2} (H-h)$
$\Rightarrow V_{2}=\frac{1}{3} \pi r^{2} \left(H-\frac{rH}{R}\right)$
$\Rightarrow \frac{dV_{2}}{dr}=\frac{1}{3}\pi H \left(2r-\frac{3r^{2}}{R}\right)$
For maxima or minima,
$\frac{dV_{2}}{dr}=0$
$\Rightarrow 2r-\frac{3r^{2}}{R}=0$
$\Rightarrow r=\frac{2R}{3}$
$\Rightarrow V_{2}=\frac{1}{3}\pi \left(\frac{2}{3}R\right)^{2} \left(H-\frac{2H}{3}\right)$
$=\frac{4\pi R^{2}H}{81}$
$V_{1}=\frac{1}{3}\pi R^{2}H $
$\therefore \frac{V_{2}}{V_{1}}=\frac{4\pi R^{2}H}{81}\div \frac{1}{3} \pi R^{2}H $
$=4 : 27 =\frac{4}{27}$